There has been 2 similar problems where I got the sign wrong so I wanted to ask you since I couldn't figure out what I was doing wrong.

The problem is: $$\lim_{x\to-\infty}\frac{\sqrt{x^2+1}}{x+1}$$

What I did was to divide both the denominator and numerator by $x$ getting: $$\lim_{x\to-\infty}\frac{\sqrt{\frac{x^2}{x^2}+\frac{1}{x^2}}}{\frac{x}{x}+\frac{1}{x}}$$ Then I thought, $x^2$s would cancel out and dividing a number by infinity would give me $0$, hence: $$\lim_{x\to-\infty}\frac{\sqrt{1+0}}{1+0}=1$$

But apparently I've made a mistake and I couldn't figure out where.

The answer is $-1$.


1 Answer 1


You must write

$$\frac{|x|\sqrt{1+\frac{1}{x^2}}}{x(1+\frac{1}{x})}$$ and


  • $\begingroup$ I'd totally forgotten that, thank you. $\endgroup$ Jun 29, 2018 at 19:03
  • $\begingroup$ Ok, i wish you a nice eveneing! $\endgroup$ Jun 29, 2018 at 19:03

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .